Sending data through AJAX to MySQL

Question

So I'm trying to send dynamic generated data with AJAX to mysql.

 <script type="text/javascript">
   var postId;

   function getdata(){
       postId = document.getElementsByTagName("post-id");
   }
   function senddata(){
       var data = getdata();

       $.ajax({
            url: "php/count_shares.php",
            type: "POST",
            data: data,
            success: function(data){
             console.log(data);
         }
     });
   }
</script>

The function is done through onClick method per picture. I'm sending a string in the post-id tag. Then with count_shares.php my code is as follows:

$opt = array(PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC
);
$server = '';
$dbname = '';
$dsn = "mysql:host=".$server.";dbname=".$dbname;
$username = '';
$password = '';

 if (isset($_POST['data'])) {

   $click = $_POST['data'];

     $sqlcs = ("UPDATE posted_ad_img SET share_count = share_count + 1 WHERE post_id = $click");

   $dbcs = new PDO($dsn, $username, $password);
   $dbcs->$opt;
   $dbcs->prepare($sqlcs);

   $dbcs->execute();
}

But nothing is being sent to my database. Any help on this matter?

Your function getdata() does not return anything, so your var data will be null. Try adding return postId; at the end of the function. — Sean
– Sean, Commented Jun 22, 2015 at 3:10
getElementsByTagName("post-id"); is invalid. post-id isn't any html element. It should be specific to html elements like input,button,select etc!! — Guruprasad J Rao
– Guruprasad J Rao, Commented Jun 22, 2015 at 3:24

Danish Bhayani · Accepted Answer · 2015-06-22 02:56:04Z

1

Try this:

$.ajax({
        url: "php/count_shares.php",
        type: "POST",
        data: "data="+data,
        success: function(data){
         console.log(data);
     }
 });

answered Jun 22, 2015 at 2:56

Danish Bhayani

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Comments

Sergey B. · Accepted Answer · 2015-06-22 14:04:24Z

0

Firstly - you don't return value from getData function. Need to change it

function getdata(){
   postId = document.getElementsByTagName("post-id");
   return postId[0].nodeValue;
}

Also you have to change your ajax request, something like this:

$.ajax({
        url: "php/count_shares.php",
        type: "POST",
        data: {data: data},
        success: function(data){
         console.log(data);
     }
 });

If you provide your html I can write more details

answered Jun 22, 2015 at 14:04

Sergey B.

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